# Acceleration vs Time Graphs

Acceleration vs time graphs tells us about an object’s velocity the same way velocity vs time graphs tell us about an object’s displacement. The change in velocity in a given time frame is equal to the area under the graph during that same time interval.

## Explaining the Acceleration-Time Graph

### Vertical Axis

The vertical axis in the graph represents the acceleration of the object. In the given graph below, reading the value of the graph at a particular time will fetch you the acceleration of the object in meters per second squared for that moment.

### Slope of the acceleration graph

The slope of the acceleration graph represents a quantity known as a jerk. Jerk is the rate of change of acceleration. In the given acceleration graph as shown below, the slope can be calculated as follows:

$slope=\frac{rise}{run}=\frac{a_{2}-a_{1}}{t_{2}-t_{1}}=\frac{\Delta a}{\Delta t}$

### Area Under the Acceleration Graph

The area under the acceleration graph represents the change in velocity. In other words, the area under the graph for a certain time interval is equal to the change in velocity during that time interval.

$Area=\Delta V$

Let us consider the below example to better understand:

The graph below shows a constant acceleration of 4 m/s2 for a time of 9 s.

Acceleration is defined as,

$\Delta a=\frac{\Delta v}{\Delta t}$

By multiplying both sides of the equation by change in time $\Delta t$, we get

$\Delta v=a\Delta t$

Substituting the values in the above equation, we get

$\Delta v=a\Delta t=(4\,m/s^2)(9\,s)=36\,m/s$

Multiplying the acceleration by the time interval is equivalent to finding the area under the curve.

The area under the curve is a rectangle. This area can be found by multiplying height and width.

The height, in this case, is 4 m/s2 and the width is 9 s.

$area=4\,m/s^2\times 9\,s=36\,m/s$

The area under any acceleration graph for a certain time interval gives the change in velocity for that time interval.

## Solved Examples

1. A race car driver is cruising at a constant velocity of 20 m/s. As she nears the finish line, the race car driver starts to accelerate. The graph shown below gives the acceleration of the race car as it starts to speed up. Assume the race car had a velocity of 20 m/s at time t=0 s. Find the final velocity of the driver when she reaches the finish line.

Solution:

We can find the change in velocity by finding the area under the acceleration graph.

$\Delta v = area=\frac{1}{2})(8\,s)(6\,m/s^2)=24\,m/s$

Substituting the values, we get

$\Delta v = area=\frac{1}{2})(8\,s)(6\,m/s^2)=24\,m/s$

This calculation gave us the change in velocity during the given time interval. To calculate the final velocity, we need to use the definition of change in velocity.

$\Delta v=v_{f}-v_i$

Substituting the values in the equation, we get

$v_{f}-20\,m/s=24\,m/s$ $v_f=44\,m/s$

Therefore, the final velocity of the racer is 44 m/s.

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